Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{n + 10}{-5n - 45} \times \dfrac{4n^2 + 44n + 72}{4n^2 + 8n - 320} $
Explanation: First factor out any common factors. $p = \dfrac{n + 10}{-5(n + 9)} \times \dfrac{4(n^2 + 11n + 18)}{4(n^2 + 2n - 80)} $ Then factor the quadratic expressions. $p = \dfrac {n + 10} {-5(n + 9)} \times \dfrac {4(n + 9)(n + 2)} {4(n + 10)(n - 8)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {(n + 10) \times 4(n + 9)(n + 2) } {-5(n + 9) \times 4(n + 10)(n - 8) } $ $p = \dfrac {4(n + 9)(n + 2)(n + 10)} {-20(n + 10)(n - 8)(n + 9)} $ Notice that $(n + 10)$ and $(n + 9)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {4(n + 9)(n + 2)\cancel{(n + 10)}} {-20\cancel{(n + 10)}(n - 8)(n + 9)} $ We are dividing by $n + 10$ , so $n + 10 \neq 0$ Therefore, $n \neq -10$ $p = \dfrac {4\cancel{(n + 9)}(n + 2)\cancel{(n + 10)}} {-20\cancel{(n + 10)}(n - 8)\cancel{(n + 9)}} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $p = \dfrac {4(n + 2)} {-20(n - 8)} $ $ p = \dfrac{-(n + 2)}{5(n - 8)}; n \neq -10; n \neq -9 $